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Does set A equal to set B ? where $$ A= \{ \emptyset, a,b,c \} $$ $$ B=\{ a,b,c \} $$

second question: I know the order of the elements does not matter in a Set. But i once read about the order of elements(or subset) matters in some case, and i can't remember what it is.

Some help from you guys would be nice. Thanks

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If one or more of $a$, $b$, $c$ is the empty set, then the two sets are the same. Otherwise they are not. –  André Nicolas Nov 25 '12 at 5:22
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3 Answers

The order of the items in the set does not matter. So these three sets are equal:

$$B = \{a,b,c\} = \{c,b,a\} = \{b,a,c\}$$

A set may have another set as a member. So a set containing the empty set and another element is distinct from a set containing just one element. For example, the $A$ and $B$ in your question are distinct sets.

The order of items in ordered pairs contained in sets matter. So, the following two sets, $C$ and $D$, are distinct because they contain different ordered pairs.

$$C = \{ (a,b) , (c,d) \} $$ $$D = \{ (b,a) , (d,c) \} $$

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"A set may have another set as a member." — actually in ZFC, a set may only have other sets as members. –  celtschk Aug 4 '13 at 10:17
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  1. Assuming $a,b,c \not = \emptyset$, we have $A-B=\lbrace \emptyset \rbrace\not =\emptyset,$ so $A\not =B$.

  2. Perhaps the time you saw that the order was of importance was when studying sequences?

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No. $A$ and $B$ are different as they have three elements in common, but $A$ also has $\emptyset$. You might think it is like adding zero, but it is not. The order matters when you have a Cartesian product or ordered $n$-tuples in general, hence the name.

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